What are the solutions to the equation x^2+7x+12=2x+8 ?

The solutions to the equation x^2+7x+12=2x+8 are x=-1,x=-4

Find the zeros of x^2+7x+12=2x+8

The zeros of x^2+7x+12=2x+8 are x=-1,x=-4

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x^{2} + 7 x + 12 = 2 x + 8

x = - 1, x = - 4

Solution steps

x^{2} + 7 x + 12 = 2 x + 8

Move

8

to the left side

x^{2} + 7 x + 4 = 2 x

Move

2 x

to the left side

x^{2} + 5 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 4 = 3

x_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 3}{2 \cdot 1}, x_{2} = \frac{- 5 - 3}{2 \cdot 1}

x = \frac{- 5 + 3}{2 \cdot 1} : - 1

x = \frac{- 5 - 3}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 1, x = - 4

\frac{x + 2}{6} = \frac{2 x - 10}{8}

\frac{1}{2} x + \frac{4}{3} x = 33

6 x^{2} + 12 x + 6

x^{2} + 9 x + 18 \geq 0

6 x^{2} - 17 x + 7

What are the solutions to the equation x^2+7x+12=2x+8 ?
The solutions to the equation x^2+7x+12=2x+8 are x=-1,x=-4
Find the zeros of x^2+7x+12=2x+8
The zeros of x^2+7x+12=2x+8 are x=-1,x=-4